The instantaneous rate of change at t = 1 for the function f (t) = is

**A. ** 2

**B. ** -1

**C. ** 0

**D. ** 9

**Answer : ****Option C**

**Explaination / Solution: **

⇒f′(1)=−e−1+e−1=0

⇒f′(1)=−e−1+e−1=0

Workspace

Report

The function f (x) = – 2 x is increasing in the interval

**A. ** x⩾1

**B. ** x≠−1
**C. ** x⩾−1

**D. ** x≠1
**Answer : ****Option A**

**Explaination / Solution: **

f(x) = x2- 2x

f'(x) = 2x - 2 = 2(x - 1)

So , f( x) is increasing if 2(x-1) 0 , i.e.if x 1

f(x) = x2- 2x

f'(x) = 2x - 2 = 2(x - 1)

So , f( x) is increasing if 2(x-1) 0 , i.e.if x 1

Workspace

Report

The function f (x) = strictly increases on

Here ,

i.e. if x(x – 2 )<0 i.e. if 0 < x < 2. Hence f is strictly increasing on [0,2]

Workspace

Report

Let g (x) be continuous in a neighbourhood of ‘a’ and g (a) ≠ 0. Let f be a function such that f ‘ (x) = g(x) then

**A. ** f is increasing at a if g (a) > 0

**B. ** none of these

**C. ** f is decreasing at a if g (a) >0

**D. ** f is increasing at a if g (a) < 0

**Answer : ****Option A**

**Explaination / Solution: **

Since g is continuous at a , therefore , if g ( a ) > 0 , then there is a neighbourhood of a, say ( a-e , a+ e ) in which g ( x ) is positive .This means that f ‘ (x)>0 in this nhd of a and hence f ( x ) is increasing at a.

Since g is continuous at a , therefore , if g ( a ) > 0 , then there is a neighbourhood of a, say ( a-e , a+ e ) in which g ( x ) is positive .This means that f ‘ (x)>0 in this nhd of a and hence f ( x ) is increasing at a.

Workspace

Report

The tangent to the parabola at the point makes with the X – axis an angle of

Therefore , slope of tangent at ( 1 , ½ ) = 1. Hence , required angle is

Workspace

Report

The equation of the tangent to the curve at the point (0, 1) is

**A. ** y + 1 = 2 x

**B. ** y – 1 = 2 x

**C. ** 1 – y = 2 x

**D. ** none of these.

**Answer : ****Option B**

**Explaination / Solution: **

Workspace

Report

Tangents to the curve at the points (1, 1) and ( – 1, 1)

therefore , slope of tangent at (1,1) = - 1 and the slope of tangent at ( - 1 ,1 )= 1 .

Now product of the slopes=1.-1= -1

Hence , the two tangents are at right angles.

Workspace

Report

The point on the curve where tangent makes an angle of with the X – axis is

**A. ** (14,12)
**B. ** (12,14)
**C. ** (2, – 2)

**D. ** (4, 2)

**Answer : ****Option A**

**Explaination / Solution: **

Workspace

Report

Let f (x) = then f (x) is
**A. ** an even function

**B. ** a decreasing function.

**C. ** an increasing function

**D. ** an odd function

**Answer : ****Option C**

**Explaination / Solution: **

Hence an increasing function.

Workspace

Report

Let f (x) = – 4x, then

**A. ** f is increasing in [−∞,1)

**B. ** f is increasing in [1,∞)

**C. ** f is decreasing in [1,∞)

**D. ** none of these.

**Answer : ****Option B**

**Explaination / Solution: **

f (x) = – 4x

f'(x) = 4(x3) - 4 = 4(x 3- 1) = 4 {(x2) + x + 1)} (x-1)

= f ‘(x) > 0,if,(x - 1) > 0,

and f ‘(x) <0, if (x -1 ) < 0.

So,f is decreasing on ( - ,1]

and f is increasing on [1,)

Workspace

Report